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I am trying to find the limit of the sequence $$s_n:=\displaystyle \prod_{i=1}^n \left(1-\frac{1}{2^i} \right)$$

The sequence is decreasing and bounded below by $0$. I guess that the limit is $0$, is there any way to show this ? Or, is there any argument which shows that the limit is not zero ?

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up vote 5 down vote accepted

The limit is definitely non-zero. Your product converges to a non-zero value, which is approximately $\approx 0.289$.

You should look up my answer here, why such infinite products can never be zero unless one of the term is itself zero. The product in your question is $\left(\dfrac12; \dfrac12\right)_{\infty}$ where $(a;q)_n$ is the Pochhammer symbol.

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In fact the limit begins with $0.2887880950866024212788997219292307800889119\cdots$

To show that the limit is finite just consider the logarithm of the product so that the term added at each stage will be of order $2^{-i}$ making the series clearly convergent.

More generally (as indicated by Marvis first) we have (Alpha) : $$\phi(z):=\prod_{i=1}^\infty 1-z^i= (z;z)_{\infty}$$

with $\phi$ the Euler function.

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